Let O be the origin and Let PQR be an arbitrary triangles. The point S is such that $\vec{O P} \cdot \vec{O Q} + \vec{O R} \cdot \vec{O S} = \vec{O R} \cdot \vec{O P} + \vec{O Q} \cdot \vec{O S} = \vec{O Q} \cdot \vec{O R} + \vec{O P} \cdot \vec{O S} Then the Triangle is P Q R has S as it's$

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in centre

Circumcentre

Orthocenter

Centroid

answer is C.

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Detailed Solution

$\begin{array}{l} \vec{O P} . (\vec{O Q} - \vec{O R}) + \vec{O S} . (\vec{O R} - \vec{O Q}) \\ \vec{O P} . \vec{R Q} + \vec{O S} . \vec{Q R} = 0 \\ \vec{R Q} . (\vec{O P} - \vec{O S}) = 0 \\ \vec{S P} . \vec{R Q} = 0 \\ ∴ \vec{S P} ⊥ \vec{R Q} \\ similarly \vec{S Q} ⊥ \vec{Q R} and \vec{S R} ⊥ \vec{P Q} \\ Therefore, orthocenter \end{array}$